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Chapter 4: Problem 57

What mass of solid aluminum hydroxide can be produced when \(50.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) is added to \(200.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{KOH} ?\)

Short Answer

Expert verified
The mass of solid aluminum hydroxide produced when $50.0 \mathrm{~mL}$ of $0.200 \mathrm{M} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}$ is added to $200.0 \mathrm{~mL}$ of $0.100 \mathrm{M} \mathrm{KOH}$ is approximately \(0.780~g\).

Step by step solution

01

1. Write the balanced chemical equation

The reaction between aluminum nitrate and potassium hydroxide produces solid aluminum hydroxide and potassium nitrate: \[ Al(NO_3)_3 + 3KOH \rightarrow Al(OH)_3(s) + 3KNO_3 \]
02

2. Calculate the moles of Al(NO3)3 and KOH

We'll calculate the moles of Al(NO_3)_3 and KOH using their volume and concentration, using the formula: \[ moles = volume (L) \times concentration (M) \] For Al(NO_3)_3: \(moles = 50.0 \times 10^{-3} L \times 0.200 M = 0.0100 mol\) For KOH: \(moles = 200.0 \times 10^{-3} L \times 0.100 M = 0.0200 mol\)
03

3. Determine the limiting reactant

To determine the limiting reactant, we'll compare the mole ratios of Al(NO_3)_3 and KOH in the balanced chemical equation. According to the equation, we need 1 mole of Al(NO_3)_3 to react with 3 moles of KOH. Mole ratio = moles of KOH / moles of Al(NO_3)_3 Mole ratio = \(0.0200 / 0.0100 = 2.00\) Since the mole ratio is less than the required 3:1, KOH is the limiting reactant.
04

4. Calculate the theoretical yield of Al(OH)3

We can use the stoichiometry from the balanced chemical equation to find the theoretical yield of Al(OH)_3: 1 mol of Al(NO_3)_3 reacts with 3 mol of KOH to produce 1 mol of Al(OH)_3 The moles of Al(OH)_3 produced will be equal to the moles of Al(NO_3)_3 available: \(moles \space of \space Al(OH)_3 = 0.0100 \space mol\)
05

5. Convert moles of Al(OH)3 to mass

We'll now convert the moles of aluminum hydroxide to its mass using its molar mass. The molar mass of Al(OH)_3 is approximately: \(1 \times Al \approx 26.98~g/mol\) \(3 \times O \approx 47.997~g/mol\) \(3 \times H \approx 3.024~g/mol\) Molar mass of Al(OH)_3 鈮� \(26.98 + 47.997 + 3.024 \approx 78.001~g/mol\) Mass of Al(OH)_3 = moles 脳 molar mass Mass of Al(OH)_3 鈮� \(0.0100~mol \times 78.001~g/mol = 0.780~g\) So, the mass of solid aluminum hydroxide produced in the reaction is approximately \(0.780~g\).

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Most popular questions from this chapter

Assign oxidation states for all atoms in each of the following compounds. a. \(\mathrm{UO}_{2}^{2+}\) f. \(\mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7}\) b. \(\mathrm{As}_{2} \mathrm{O}_{3} \quad\) g. \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) c. \(\mathrm{NaBiO}_{3}\) h. \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\) d. As i. \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) e. \(\mathrm{HAsO}_{2}\)

Balance each of the following oxidation-reduction reactions by using the oxidation states method. a. \(\mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) b. \(\mathrm{Mg}(s)+\mathrm{HCl}(a q) \rightarrow \mathrm{Mg}^{2+}(a q)+\mathrm{Cl}^{-}(a q)+\mathrm{H}_{2}(g)\) c. \(\mathrm{Co}^{3+}(a q)+\mathrm{Ni}(s) \rightarrow \mathrm{Co}^{2+}(a q)+\mathrm{Ni}^{2+}(a q)\) d. \(\mathrm{Zn}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{ZnSO}_{4}(a q)+\mathrm{H}_{2}(g)\)

Write the balanced formula equation for the acid-base reactions that occur when the following are mixed. a. potassium hydroxide (aqueous) and nitric acid b. barium hydroxide (aqueous) and hydrochloric acid c. perchloric acid \(\left[\mathrm{HClO}_{4}(a q)\right]\) and solid iron(III) hydroxide d. solid silver hydroxide and hydrobromic acid e. aqueous strontium hydroxide and hydroiodic acid

You made \(100.0 \mathrm{~mL}\) of a lead(II) nitrate solution for lab but forgot to cap it. The next lab session you noticed that there was only \(80.0 \mathrm{~mL}\) left (the rest had evaporated). In addition, you forgot the initial concentration of the solution. You decide to take \(2.00 \mathrm{~mL}\) of the solution and add an excess of a concentrated sodium chloride solution. You obtain a solid with a mass of \(3.407 \mathrm{~g}\). What was the concentration of the original lead(II) nitrate solution?

On the basis of the general solubility rules given in Table 4.1, predict which of the following substances are likely to be soluble in water. a. aluminum nitrate b. magnesium chloride c. rubidium sulfate d. nickel(II) hydroxide e. lead(II) sulfide f. magnesium hydroxide g. iron(III) phosphate
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